Answer:
Earth: 22.246 N
Moon: 3.71 N
Jupiter: 58.72 N
Explanation:
The mass of an object will remain constant in any location, its weight however, can fluctuate depending on its location. For example, a golf ball will weigh less on the moon, but its mass will not be different if it was on earth.
To calculate anything, we need to convert to standard measurements.
5.00 lbs = 2.27 kg
On earth, gravity is measured to be 9.8 m/s², so the weight in Newtons on Earth would be: (2.27 kg) x (9.8 m/s²) = 22.246 N
Repeated on the moon where gravity is (9.8 m/s²) x (1/6) = 1.633 m/s², so the weight in Newtons on the moon would be: (2.27 kg) x (1.633 m/s²) = 3.71 N
Repeated on Jupiter where gravity is (9.8 m/s²) x (2.64) = 25.87 m/s², so the wight in Newtons on Jupiter would be: (2.27 kg) x (25.87 m/s²) = 58.72 N
A.
Destructive interference is when two waves cancel each other out or when the crest of one wave passes through the trough of another wave.
The total work done on the car is 784Joule.
<h3>What's the acceleration of the car?</h3>
- As per Newton's equation of motion, V= U+at
- U= initial velocity= 0 m/s
V= vinal velocity= 20m/s
t= time = 10s
a= acceleration
=> a= 20/10= 2m/s²
<h3>What's the distance covered by the car in 10 seconds?</h3>
- As per Newton's equation of motion,
V²-U² = 2aS
- S= distance covered by the car
- So, 20²-0=2×2×S=4S
=> 400= 4S
=> S= 400/4= 100m
<h3>What's the work done on the car due to frictional force?</h3>
Work done by frictional force= frictional force × distance
= (0.2×4×9.8)×100
= 784Joule
Thus, we can conclude that the work done on the car is 784Joule.
Learn more about the work done here:
brainly.com/question/25573309
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<u>Answer :</u>
(a) d = 0.25 m
(b) d = 0.5 m
<u>Explanation :</u>
It is given that,
Frequency of sound waves, f = 686 Hz
Speed of sound wave at 
is, v = 343 m/s
(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.
........(1)
Velocity of sound wave is given by :
Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.
(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.
( n = integers )
Let n = 1
So, 
Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.
Answer:
5
Explanation:
Number of habitable planets = 1000
Fraction of planet with life = 1/10
Fraction of planet with life and civilization (before) = 1/4
Fraction of planet with life and civilization (now) =1/5
Therefore multiplying we have:
1000×1/10×1/4×1/5 = 5
